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The easiest way to find a continued fraction for $\sqrt{n}$ is to use a generalized continued fraction (continued fractions where the numerator is something besides 1). Starting from $\sqrt{n}$, partition ${n}$ as the sum of the perfect square which is less than or equal to ${n}$, ${s^{2}}$, and a remainder, ${r}$, like this: $\sqrt{s^{2} + r}$. The ...


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If your $n$ is not a perfect square, I'll show it with an example. Consider $\sqrt{23}.$ $$\sqrt{23}\approx4.8$$ Here I denote successive irrational by $x_k$ and it's integer part by $a_k.$ $$x_0=\sqrt{23}=4+(\sqrt{23}-4)\rightarrow \,\ a_0=4$$ $$x_1=\dfrac{1}{x_0-a_0}=\dfrac{1}{\sqrt{23}-4}=\dfrac{\sqrt{23}+4}{7}=1+\dfrac{\sqrt{23}-3}{7}\rightarrow ...


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Let's try, for example, $\sqrt5$: Since that $2\lt\sqrt5\lt3$, the first term is $2$. Subtract $2$ and invert: $$ \frac1{\sqrt5-2}=\sqrt5+2 $$ Since $4\le\sqrt5+2\le5$, the next term is $4$. Subtract $4$ and invert: $$ \frac1{\sqrt5+2-4}=\sqrt5+2 $$ We are at the same point as the previous line. Thus, the continued fraction is $$ (2;4,4,4,4,\dots) $$ Let's ...


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The key is to find reciprocals, using conjugates. For example, the reciprocal of $\sqrt{7}-2$ is $\frac{\sqrt{7}+2}3$, which you can check by multiplying those two together. To start with, $$\sqrt{7}=2+(\sqrt{7}-2)\\=2+\frac{1}{\frac{\sqrt{7}+2}3}\\=2+\frac{1}{1+\frac{\sqrt{7}-1}3}$$ Carry on like that. You will only need to know that $\sqrt{7}$ is ...


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Suppose $f$ continuous at $a$ and let $(x_n)$ a sequence such that $x_n \to a$ Let $\varepsilon > 0$. By continuity of $f$ at $a$ there exists $\alpha > 0$ such that : $$|x-a| < \alpha \Rightarrow |f(x)-f(a) | < \varepsilon $$ Since $(x_n) \to a$ there exists $N$ integer such that : $$n \geq N \Rightarrow |x_n-a| < \alpha$$ Now it's ...


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If $$z_1 = 1 + 2/(3 + 4/(5 + 6/( ... = {1\over \exp(1/2)-1 }$$ (as given in the answer before) and $$z_3 =3 + 4/(5 + 6/( ... $$ then $$ z_1 = 1 + 2/z_3$$ or $$z_3 = { 2 \over z_1-1} = { 2 \over {1 \over \exp(1/2)-1}-1} = { 2 \exp(1/2)-2 \over 2 - \exp(1/2) } $$ From this $z_5,z_7,...$ follow analoguously, and they are all rational compositions of the ...


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Well, I came up with a much better solution than my previous one, but it's too substantial to add to my old answer. We are looking for irrational numbers $x$ with convergents $\frac{p_n}{q_n}$ such that, if we define $f_x(n)=p_n-q_nx$, there exists some real number $c$ such that $f_x(n+1)=cf_x(n)$. Notice that, according to Wikipedia, if $x$ has continued ...


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This is a really awesome observation of yours. The main thing that will help you here is to think in purely algebraic terms. Notably, any product, sum, quotient, etc. involving only rational numbers and $\sqrt{k}$ can be written in the form $a+b\sqrt{k}$. In particular, if we define $\phi_k=\frac{\sqrt{k^2+4}-k}2$, notice that this is a root of ...


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I got $\displaystyle \sum_{3n+2}^\infty \phi^m = \frac{\phi^{3n+2}}{1- \phi}$ and then I could not find a good way to simplify $1 - \phi$: $$ 1 - \phi = \frac{3-\sqrt{5}}{2}$$ To get you started, use the identity $\phi^2 = -\phi + 1$, since you have written $\phi = \tfrac{\sqrt{5}-1}{2}$. And say: $$ \frac{1}{1-\phi} = \frac{1+\phi}{1 - \phi^2} = ...


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The opposite inequality is true. Since $p_n=a_np_{n-1}+p_{n-2}$ and $q_n=a_nq_{n-1}+q_{n-2}$, we have $$ p_n-\alpha q_n = a_n(p_{n-1}-\alpha q_{n-1}) + (p_{n-2}-\alpha q_{n-2}). $$ Since consecutive convergents alternate sides around $\alpha$, it follows that $$ \theta_n=-a_n\theta_{n-1}+\theta_{n-2}. $$ Finally, since the $\theta_n$'s are nonnegative and ...



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